A rocket is launched normal to the surface of the Earth

PHYSICS · Single correct · CBSE · Physics · Gravitation · Gravitation

Problem

A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is 3×1053\times 10^5 times heavier than the Earth and is at a distance 2.5×1042.5\times 10^4 times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is ve=11.2 km s−1v_e = 11.2\,\text{km s}^{-1}. The minimum initial velocity (vsv_s) required for the rocket to be able to leave the Sun-Earth system is closest to. (Ignore the rotation and revolution of the Earth and the presence of any other planet) (A) vs=22 km s−1v_s = 22\,\text{km s}^{-1} (B) vs=72 km s−1v_s = 72\,\text{km s}^{-1} (C) vs=42 km s−1v_s = 42\,\text{km s}^{-1} (D) vs=62 km s−1v_s = 62\,\text{km s}^{-1}

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